Measured by one unit, then the square of the number expressing the hypotenuse equal to the sum squares of numbers, height pressing the legs.
This theorem is usually expressed abbreviated as follows:
The square of the hypotenuse is equal to the sum of the squares of the legs.
This relationship was first noticed by the Greek geometer Pythagoras (VI century BC) and therefore bears his name - Pythagorean theorem .
Theorem.
acute angle, equal to the sum of the squares of the other two sides without twice the product of any of these sides by its segment from the vertex acute angle up to height.
Let BWITH- side of the triangle ABWITH(Fig. 1 and Fig. 2), lying opposite the acute angle A, And BD- height lowered on any of the other sides, for example, on AWITH(or its continuation). It is required to prove that:
B.C. 2 = AB 2 + AC 2 - 2AS. AD.
From right triangles BDC And ABD we output:
BC 2= BD 2+DWITH 2 [ 1 ] ;
BD 2= AB 2 - AD 2 [ 2] .
On the other side: DWITH= AC-AD(Fig. 1) or DWITH= AD-AS(Fig. 2). In both cases for DWITH 2 we get the same expression:
DWITH 2 = (AWITH-AD) 2 = AWITH 2 - 2AWITH . AD + AD 2 ;
DWITH 2 = (AD-AWITH) 2 = AD 2 - 2AD . AWITH + AWITH 2 .
Substituting into equality instead BD 2 And DC 2 their expressions from the equalities and , we obtain:
BC 2= AB 2 - A D 2 + AWITH 2 - 2 AWITH . AD + A D 2 .
This is equality, after the reduction of members -AD 2 And + AD 2 , and is the very thing that needed to be proven.
Comment. The proven theorem remains true even when the angle WITH straight. Then the segment CD will go to zero, i.e. AC will become equal to AD, and we will have:
BC 2= AB 2+ AWITH 2 - 2AWITH 2 = AB 2 - AWITH 2 .
Which is consistent with the theorem about squared hypotenuse.
Theorem.
In a triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides, added to twice the product of any of these sides by the segment of its continuation from the vertex of the obtuse angle to the altitude. The proof is similar to the previous one.
Consequence.
From the last three theorems we deduce that the square of a side of a triangle is equal to, less than, or more than the amount squares of other sides, depending on whether the opposite angle is right, acute or obtuse.
this implies reverse offer: The angle of the triangle will be right, acute or obtuse, depending on whether it is a square the opposite side equal to, less than, or greater than the sum of the squares of the other sides.
Calculating the height of a triangle based on its sides.
Let us denote the height dropped to side a of the triangle ABWITH, through h a. To calculate it, first from the equation:
b 2 = a 2 + from 2 - 2aWith’ .
find the base segment c’:
.
Then from DABD we determine the height as a leg:
.
In the same way, you can determine the heights h b and h c, lowered to sides b and c.
Calculation of the medians of a triangle based on its sides.
Let the sides of the triangle be given ABWITH and you need to calculate its median BD. To do this, let’s extend it to a distance DE = BD and period E connect with A And WITH. Then we get a parallelogram ABCE.
Applying the previous theorem to it, we find: BE 2 = 2 AB 2 + 2 BC 2 -AC 2.
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As a rule, two triangles are considered similar if they have same shape, even if they are different sizes, rotated or even upside down.
The mathematical representation of two similar triangles A 1 B 1 C 1 and A 2 B 2 C 2 shown in the figure is written as follows:
ΔA 1 B 1 C 1 ~ ΔA 2 B 2 C 2
Two triangles are similar if:
1. Each angle of one triangle is equal to the corresponding angle of another triangle:
∠A 1 = ∠A 2 , ∠B 1 = ∠B 2 And ∠C 1 = ∠C 2
2. The ratios of the sides of one triangle to the corresponding sides of another triangle are equal to each other:
$\frac(A_1B_1)(A_2B_2)=\frac(A_1C_1)(A_2C_2)=\frac(B_1C_1)(B_2C_2)$
3. Relationships two sides one triangle to the corresponding sides of another triangle are equal to each other and at the same time
the angles between these sides are equal:
$\frac(B_1A_1)(B_2A_2)=\frac(A_1C_1)(A_2C_2)$ and $\angle A_1 = \angle A_2$
or
$\frac(A_1B_1)(A_2B_2)=\frac(B_1C_1)(B_2C_2)$ and $\angle B_1 = \angle B_2$
or
$\frac(B_1C_1)(B_2C_2)=\frac(C_1A_1)(C_2A_2)$ and $\angle C_1 = \angle C_2$
Do not confuse similar triangles with equal triangles. Equal triangles have equal corresponding side lengths. Therefore, for congruent triangles:
$\frac(A_1B_1)(A_2B_2)=\frac(A_1C_1)(A_2C_2)=\frac(B_1C_1)(B_2C_2)=1$
It follows from this that everything equal triangles are similar. However, not all similar triangles are equal.
Although the above notation shows that to find out whether two triangles are similar or not, we must know the values of the three angles or the lengths of the three sides of each triangle, to solve problems with similar triangles it is enough to know any three of the values mentioned above for each triangle. These quantities can be in various combinations:
1) three angles of each triangle (you don’t need to know the lengths of the sides of the triangles).
Or at least 2 angles of one triangle must be equal to 2 angles of another triangle.
Since if 2 angles are equal, then the third angle will also be equal. (The value of the third angle is 180 - angle1 - angle2)
2) the lengths of the sides of each triangle (you don’t need to know the angles);
3) the lengths of the two sides and the angle between them.
Next we will look at solving some problems with similar triangles. First we will look at the problems that can be solved direct use above rules, and then discuss some practical problems, which are solved using the method of similar triangles.
Practice problems with similar triangles
Example #1:
Show that the two triangles in the figure below are similar. 
Solution:
Since the lengths of the sides of both triangles are known, the second rule can be applied here:
$\frac(PQ)(AB)=\frac(6)(2)=3$ $\frac(QR)(CB)=\frac(12)(4)=3$ $\frac(PR)(AC )=\frac(15)(5)=3$
Example #2:
Show that two given triangles are similar and determine the lengths of the sides PQ And PR.
Solution:
∠A = ∠P And ∠B = ∠Q, ∠C = ∠R(since ∠C = 180 - ∠A - ∠B and ∠R = 180 - ∠P - ∠Q)
It follows from this that the triangles ΔABC and ΔPQR are similar. Hence:
$\frac(AB)(PQ)=\frac(BC)(QR)=\frac(AC)(PR)$
$\frac(BC)(QR)=\frac(6)(12)=\frac(AB)(PQ)=\frac(4)(PQ) \Rightarrow PQ=\frac(4\times12)(6) = 8$ and
$\frac(BC)(QR)=\frac(6)(12)=\frac(AC)(PR)=\frac(7)(PR) \Rightarrow PR=\frac(7\times12)(6) = 14$
Example #3:
Determine the length AB in this triangle. 
Solution:
∠ABC = ∠ADE, ∠ACB = ∠AED And ∠A general => triangles ΔABC And ΔADE are similar.
$\frac(BC)(DE) = \frac(3)(6) = \frac(AB)(AD) = \frac(AB)(AB + BD) = \frac(AB)(AB + 4) = \frac(1)(2) \Rightarrow 2\times AB = AB + 4 \Rightarrow AB = 4$
Example #4:
Determine length AD(x) geometric figure on the image. 
Triangles ΔABC and ΔCDE are similar because AB || DE and they have one in common top corner C.
We see that one triangle is a scaled version of the other. However, we need to prove this mathematically.
AB || DE, CD || AC and BC || E.C.
∠BAC = ∠EDC and ∠ABC = ∠DEC
Based on the above and taking into account the availability total angle C, we can claim that triangles ΔABC and ΔCDE are similar.
Hence:
$\frac(DE)(AB) = \frac(7)(11) = \frac(CD)(CA) = \frac(15)(CA) \Rightarrow CA = \frac(15 \times 11)(7 ) = 23.57$
x = AC - DC = 23.57 - 15 = 8.57
Practical examples
Example #5:
The factory uses an inclined conveyor belt to transport products from level 1 to level 2, which is 3 meters higher than level 1, as shown in the figure. The inclined conveyor is serviced from one end to level 1 and from the other end to a workplace located at a distance of 8 meters from the level 1 operating point. 
The factory wants to upgrade the conveyor to access the new level, which is 9 meters above level 1, while maintaining the conveyor's inclination angle.
Determine the distance at which the new work station must be installed to ensure that the conveyor will operate at its new end at level 2. Also calculate the additional distance the product will travel when moving to the new level.
Solution:
First, let's label each intersection point with a specific letter, as shown in the figure.
Based on the reasoning given above in the previous examples, we can conclude that the triangles ΔABC and ΔADE are similar. Hence,
$\frac(DE)(BC) = \frac(3)(9) = \frac(AD)(AB) = \frac(8)(AB) \Rightarrow AB = \frac(8 \times 9)(3 ) = 24 m$
x = AB - 8 = 24 - 8 = 16 m
Thus, new item should be installed at a distance of 16 meters from an existing point.
And since the structure consists of right triangles, we can calculate the distance of movement of the product as follows:
$AE = \sqrt(AD^2 + DE^2) = \sqrt(8^2 + 3^2) = 8.54 m$
Similarly, $AC = \sqrt(AB^2 + BC^2) = \sqrt(24^2 + 9^2) = 25.63 m$
which is the distance that the product travels in this moment upon reaching the existing level.
y = AC - AE = 25.63 - 8.54 = 17.09 m
this is the additional distance that the product must travel to reach a new level.
Example #6:
Steve wants to visit his friend who recently moved to a new house. Road map directions to the house of Steve and his friend, along with the distances known to Steve, are shown in the figure. Help Steve get to his friend's house in the shortest possible way. 
Solution:
A road map can be represented geometrically in the following form, as it shown on the picture. 
We see that triangles ΔABC and ΔCDE are similar, therefore:
$\frac(AB)(DE) = \frac(BC)(CD) = \frac(AC)(CE)$
The problem statement states that:
AB = 15 km, AC = 13.13 km, CD = 4.41 km and DE = 5 km
Using this information we can calculate the following distances:
$BC = \frac(AB \times CD)(DE) = \frac(15 \times 4.41)(5) = 13.23 km$
$CE = \frac(AC \times CD)(BC) = \frac(13.13 \times 4.41)(13.23) = 4.38 km$
Steve can get to his friend's house using the following routes:
A -> B -> C -> E -> G, total distance is 7.5+13.23+4.38+2.5=27.61 km
F -> B -> C -> D -> G, total distance is 7.5+13.23+4.41+2.5=27.64 km
F -> A -> C -> E -> G, total distance is 7.5+13.13+4.38+2.5=27.51 km
F -> A -> C -> D -> G, total distance is 7.5+13.13+4.41+2.5=27.54 km
Therefore, route No. 3 is the shortest and can be offered to Steve.
Example 7:
Trisha wants to measure the height of the house, but she doesn't have it the right tools. She noticed that there was a tree growing in front of the house and decided to use her resourcefulness and knowledge of geometry acquired at school to determine the height of the building. She measured the distance from the tree to the house, the result was 30 m. She then stood in front of the tree and began to move back until the top edge of the building became visible above the top of the tree. Trisha marked this place and measured the distance from it to the tree. This distance was 5 m.
The height of the tree is 2.8 m, and the height of Trisha's eye level is 1.6 m. Help Trisha determine the height of the building. 
Solution:
The geometric representation of the problem is shown in the figure. 
First we use the similarity of triangles ΔABC and ΔADE.
$\frac(BC)(DE) = \frac(1.6)(2.8) = \frac(AC)(AE) = \frac(AC)(5 + AC) \Rightarrow 2.8 \times AC = 1.6 \times (5 + AC) = 8 + 1.6 \times AC$
$(2.8 - 1.6) \times AC = 8 \Rightarrow AC = \frac(8)(1.2) = 6.67$
We can then use the similarity of triangles ΔACB and ΔAFG or ΔADE and ΔAFG. Let's choose the first option.
$\frac(BC)(FG) = \frac(1.6)(H) = \frac(AC)(AG) = \frac(6.67)(6.67 + 5 + 30) = 0.16 \Rightarrow H = \frac(1.6 )(0.16) = 10 m$
228. In this chapter we will mainly understand by the designations of segments AB, AC, etc., the numbers expressing them.
We know (item 226) that if two segments a and b are given geometrically, then we can construct an average proportional between them. Let now the segments be given not geometrically, but by numbers, i.e. by a and b we mean numbers expressing 2 given segments. Then finding the average proportional segment will come down to finding the number x from the proportion a/x = x/b, where a, b and x are numbers. From this proportion we have:
x 2 = ab
x = √ab
229. Let us have a right triangle ABC (drawing 224).
Let's lower it from the top right angle(∠B straight line) perpendicular BD to the hypotenuse AC. Then from paragraph 225 we know:
1) AC/AB = AB/AD and 2) AC/BC = BC/DC.
From here we get:
AB 2 = AC AD and BC 2 = AC DC.
Adding the resulting equalities piece by piece, we get:
AB 2 + BC 2 = AC AD + AC DC = AC(AD + DC).
i.e. the square of the number expressing the hypotenuse is equal to the sum of the squares of the numbers expressing the legs right triangle .
In short they say: The square of the hypotenuse of a right triangle is equal to the sum of the squares of the legs.
If we give the resulting formula a geometric interpretation, we will obtain the Pythagorean theorem already known to us (item 161):
a square built on the hypotenuse of a right triangle is equal to the sum of the squares built on the legs.
From the equation AB 2 + BC 2 = AC 2, sometimes you have to find a leg of a right triangle, using the hypotenuse and another leg. We get, for example:
AB 2 = AC 2 – BC 2 and so on
230. Found numerical ratio between the sides of a right triangle allows you to solve many computational problems. Let's solve some of them:
1. Calculate area equilateral triangle on this side of it.
Let ∆ABC (drawing 225) be equilateral and each side expressed by a number a (AB = BC = AC = a). To calculate the area of this triangle, you must first find out its height BD, which we will call h. We know that in an equilateral triangle, the height BD bisects the base AC, i.e. AD = DC = a/2. Therefore, from the right triangle DBC we have:
BD 2 = BC 2 – DC 2,
h 2 = a 2 – a 2 /4 = 3a 2 /4 (perform subtraction).
From here we have:
(we take out the multiplier from under the root).
Therefore, calling the number expressing the area of our triangle in terms of Q and knowing that the area ∆ABC = (AC BD)/2, we find:
We can look at this formula as one of the ways to measure the area of an equilateral triangle: we need to measure its side in linear units, square the found number, multiply the resulting number by √3 and divide by 4 - we get the expression for the area in square (corresponding) units.
2. The sides of the triangle are 10, 17 and 21 lines. unit Calculate its area.
Let us lower the height h in our triangle (drawing 226) to the larger side - it will certainly pass inside the triangle, since in the triangle obtuse angle can only be positioned against larger side. Then the larger side, = 21, will be divided into 2 segments, one of which we denote by x (see drawing) - then the other = 21 – x. We get two right triangles, from which we have:
h 2 = 10 2 – x 2 and h 2 = 17 2 – (21 – x) 2
Since the left sides of these equations are the same, then
10 2 – x 2 = 17 2 – (21 – x) 2
Carrying out the actions we get:
10 2 – x 2 = 289 – 441 + 42x – x 2
Simplifying this equation, we find:
Then from the equation h 2 = 10 2 – x 2, we get:
h 2 = 10 2 – 6 2 = 64
and therefore
Then the required area will be found:
Q = (21 8)/2 sq. unit = 84 sq. unit
3. You can solve a general problem:
how to calculate the area of a triangle based on its sides?
Let the sides of triangle ABC be expressed by the numbers BC = a, AC = b and AB = c (drawing 227). Let us assume that AC is the larger side; then the height BD will go inside ∆ABC. Let's call: BD = h, DC = x and then AD = b – x.
From ∆BDC we have: h 2 = a 2 – x 2 .
From ∆ABD we have: h 2 = c 2 – (b – x) 2,
from where a 2 – x 2 = c 2 – (b – x) 2.
Solving this equation, we consistently obtain:
2bx = a 2 + b 2 – c 2 and x = (a 2 + b 2 – c 2)/2b.
(The latter is written on the basis that the numerator 4a 2 b 2 – (a 2 + b 2 – c 2) 2 can be considered as an equality of squares, which we decompose into the product of the sum and the difference).
This formula is transformed by introducing the perimeter of the triangle, which we denote by 2p, i.e.
Subtracting 2c from both sides of the equality, we get:
a + b + c – 2c = 2p – 2c or a + b – c = 2(p – c):
We will also find:
c + a – b = 2(p – b) and c – a + b = 2(p – a).
Then we get:
(p expresses the semi-perimeter of the triangle).
This formula can be used to calculate the area of a triangle based on its three sides.
231. Exercises.
232. In paragraph 229 we found the relationship between the sides of a right triangle. You can find a similar relationship for the sides (with the addition of another segment) of an oblique triangle.
Let us first have ∆ABC (drawing 228) such that ∠A is acute. Let's try to find an expression for the square of side BC lying opposite this acute angle (similar to how in paragraph 229 we found the expression for the square of the hypotenuse).
By constructing BD ⊥ AC, we obtain from the right triangle BDC:
BC 2 = BD 2 + DC 2
Let's replace BD2 by defining it from ABD, from which we have:
BD 2 = AB 2 – AD 2,
and replace the segment DC through AC – AD (obviously, DC = AC – AD). Then we get:
BC 2 = AB 2 – AD 2 + (AC – AD) 2 = AB 2 – AD 2 + AC 2 – 2AC AD + AD 2
Having reduced similar terms, we find:
BC 2 = AB 2 + AC 2 – 2AC AD.
This formula reads: the square of the side of a triangle opposite the acute angle is equal to the sum of the squares of its two other sides, minus twice the product of one of these sides by its segment from the vertex of the acute angle to the height.
233. Now let ∠A and ∆ABC (drawing 229) be obtuse. Let us find an expression for the square of the side BC lying opposite the obtuse angle.
Having constructed the height BD, it will now be located slightly differently: at 228 where ∠A is acute, points D and C are located on one side of A, and here, where ∠A is obtuse, points D and C will be located along different sides from A. Then from the rectangular ∆BDC we obtain:
BC 2 = BD 2 + DC 2
We can replace BD2 by defining it from the rectangular ∆BDA:
BD 2 = AB 2 – AD 2,
and the segment DC = AC + AD, which is obvious. Replacing, we get:
BC 2 = AB 2 – AD 2 + (AC + AD) 2 = AB 2 – AD 2 + AC 2 + 2AC AD + AD 2
Carrying out the reduction of similar terms we find:
BC 2 = AB 2 + AC 2 + 2AC AD,
i.e. the square of the side of a triangle lying opposite the obtuse angle is equal to the sum of the squares of its two other sides, plus twice the product of one of them by its segment from the vertex of the obtuse angle to the height.
This formula, as well as the formula of paragraph 232, admit of a geometric interpretation, which is easy to find.
234. Using the properties of paragraphs. 229, 232, 233, we can, if given the sides of a triangle in numbers, find out whether the triangle has a right angle or an obtuse angle.
A right or obtuse angle in a triangle can only be located opposite the larger side; what is the angle opposite it is easy to find out: this angle is acute, right or obtuse, depending on whether the square of the larger side is less than, equal to or greater than the sum of the squares of the other two sides .
Find out whether the following triangles, defined by their sides, have a right or an obtuse angle:
1) 15 dm., 13 dm. and 14 in.; 2) 20, 29 and 21; 3) 11, 8 and 13; 4) 7, 11 and 15.
235. Let us have parallelogram ABCD(draft 230); Let us construct its diagonals AC and BD and its altitudes BK ⊥ AD and CL ⊥ AD.
Then, if ∠A (∠BAD) is sharp, then ∠D (∠ADC) is certainly obtuse (since their sum = 2d). From ∆ABD, where ∠A is considered acute, we have:
BD 2 = AB 2 + AD 2 – 2AD AK,
and from ∆ACD, where ∠D is obtuse, we have:
AC 2 = AD 2 + CD 2 + 2AD DL.
In the last formula, let us replace the segment AD with the segment BC equal to it and DL with the segment AK equal to it (DL = AK, because ∆ABK = ∆DCL, which is easy to see). Then we get:
AC2 = BC2 + CD2 + 2AD · AK.
Adding the expression for BD2 with last expression for AC 2, we find:
BD 2 + AC 2 = AB 2 + AD 2 + BC 2 + CD 2,
since the terms –2AD · AK and +2AD · AK cancel each other out. We can read the resulting equality:
The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
236. Calculating the median and bisector of a triangle from its sides. Let in triangle ABC(drawing 231) the median BM is constructed (i.e. AM = MC). Knowing the sides ∆ABC: BC = a, AC = b and AB = c, calculate the median BM.
Let's continue BM and set aside the segment MD = BM. By connecting D with A and D with C, we get parallelogram ABCD (this is easy to figure out, since ∆AMD = ∆BMC and ∆AMB = ∆DMC).
Calling the median BM in terms of m, we get BD = 2m and then, using the previous paragraph, we have:
237. Calculation of the radius circumscribed about a triangle of a circle. Let a circle O be described around ∆ABC (drawing 233). Let us construct the diameter of the circle BD, the chord AD and the height of the triangle BH.
Then ∆ABD ~ ∆BCH (∠A = ∠H = d - angle A is a right angle, because it is inscribed, based on the diameter BD and ∠D = ∠C, as inscribed, based on one arc AB). Therefore we have:
or, calling the radius OB by R, the height BH by h, and the sides AB and BC, as before, respectively by c and a:
but area ∆ABC = Q = bh/2, whence h = 2Q/b.
Therefore, R = (abc) / (4Q).
We can (item 230 of problem 3) calculate the area of triangle Q based on its sides. From here we can calculate R from the three sides of the triangle.
238. Calculation of the radius of a circle inscribed in a triangle. Let us write in ∆ABC, the sides of which are given (drawing 234), a circle O. Connecting its center O with the vertices of the triangle and with the tangent points D, E and F of the sides to the circle, we find that the radii of the circle OD, OE and OF serve as the altitudes of the triangles BOC, COA and AOB.
Calling the radius of the inscribed circle through r, we have:
